Question: 33. A wire of resistance R is cut into 8 equal pieces. From these pieces two equivalent resistances are made by adding four of these together in parallel. Then these two sets are added in series. The net effective resistance of the combination is:
(1) \frac{R}{64}
(2) \frac{R}{32}
(3) \frac{R}{16}
(4) \frac{R}{8}
Answer: Option (3)
Explanation:
When a uniform wire of resistance R is cut into 8 equal pieces,
the length of each piece becomes \frac{1}{8}
of the original wire. Since resistance is directly proportional to length,
the resistance of each piece is:
\frac{R}{8}Now, four such pieces are connected in parallel to form one equivalent resistance.
Let the equivalent resistance of one parallel set be R_{p}.
For four equal resistances \frac{R}{8} connected in parallel:
\frac{1}{R_{p}} = 4 \times \frac{1}{\frac{R}{8}} \frac{1}{R_{p}} = \frac{32}{R}So,
R_{p} = \frac{R}{32}Two such identical parallel combinations are then connected in series.
For series combination, resistances add directly:
R_{\text{eq}} = R_{p} + R_{p} = 2 \times \frac{R}{32}Simplifying,
R_{\text{eq}} = \frac{R}{16}Hence, the net effective resistance of the combination is \frac{R}{16},
which corresponds to Option (3).